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Oksana_A [137]
3 years ago
8

A sample of pure alumina hydrate was obtained. A 5.000 g sample of the material was heated carefully in a vacuum oven until no m

ore mass was lost from the sample. After heating, the final weight of the material was 2.6763 g. What was the formula of the hydrated alumina, Al2O3•xH2O? (Enter a whole number for "x") (mol. wt. Al2O3 = 101.96)
Chemistry
1 answer:
lapo4ka [179]3 years ago
6 0

Answer: The formula of the hydrated alumina is  Al_2O_3.5H_2O

Explanation:

Decomposition of hydrated alumina is given by:

Al_2O_3.xH_2O\rightarrow Al_2O_3+xH_2O


Molar mass of Al_2O_3 = 101.96 g/mol

According to stoichiometry:

(101.96+18x) g of  Al_2O_3.xH_2O decomposes to give 101.96 g of

Al_2O_3

Thus 5.000 g of   Al_2O_3.xH_2O decomposes to give=\frac {101.96}{(101.96+18x)}\times 5.000 of H_2O

But it is given 5.000 g of a sample of hydrated salt Al_2O_3.xH_2O was found to contain 2.6763 g of unhydrated salt

Thus we can equate the two equations:

\frac{101.96}{(101.96+18x)}\times 5.000=2.6763

x=5

Thus the formula of the hydrated alumina is  Al_2O_3.5H_2O

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